Question: Consider the parametric curve: $\begin{aligned} x&=7e^t \\\\ y&=3t^{5} \end{aligned}$ Which integral gives the arc length of the curve over the interval from $t=1$ to $t=5$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\int_{1}^{5} \sqrt{49e^{t^2}+225t^{16}}\,dt$ (Choice B) B $\int_{1}^{5} \sqrt{7e^t+3t^5}\,dt$ (Choice C) C $\int_{1}^{5} \sqrt{49e^{2t}+225t^8}\,dt$ (Choice D) D $\int_{1}^{5} \sqrt{49e^{2t}+9t^{10}}\,dt$
Solution: This is the formula for the arc length of a parametric curve over the interval $[a, b]$ : $\int_a^b\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\, dt$ [Where does this formula come from?] Let's find $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ : $\begin{aligned} \dfrac{dx}{dt}&=\dfrac{d}{dt}\left[7e^t\right] \\\\ &=7e^t \\\\\\ \dfrac{dy}{dt}&=\dfrac{d}{dt}\left[3t^{5}\right] \\\\ &=15t^4 \end{aligned}$ Now we can find the expression for the arc length: $\begin{aligned} &\phantom{=}\int_{1}^{5} \sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt \\\\ &=\int_{1}^{5} \sqrt{\left(7e^t\right)^2+\left(15t^4\right)^2}\,dt \\\\ &=\int_{1}^{5} \sqrt{49e^{2t}+225t^8}\,dt \end{aligned}$ In conclusion, this integral gives the arc length of the curve over the interval from $t=1$ to $t=5$ : $\int_{1}^{5} \sqrt{49e^{2t}+225t^8}\,dt$